Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2
Explanation:
(1) Given mass = 0.125 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2
(2) Given mass = 0.250 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2
(3) Given mass = 0.375 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2
(4) Given mass = 0.500 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
Explanation:
It is given that,
Mass of bumper car, m₁ = 202 kg
Initial speed of the bumper car, u₁ = 8.5 m/s
Mass of the other car, m₂ = 355 kg
Initial velocity of the other car is 0 as it at rest, u₂ = 0
Final velocity of the other car after collision, v₂ = 5.8 m/s
Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁
Using the conservation of linear momentum as :


p₁ = m₁v₁ = -342 kg-m/s
So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.
Answer:
The value of the linear coefficient of thermal expansion is : α=1.01 *10⁻⁵ (ºC)⁻¹
Explanation:
Li = 0.2m
ΔL = 0.2 mm = 0.0002m
T1 = 21ºC
T2 = 120ºC
ΔT =99ºC
α =ΔL/(Li*ΔT)
α =0.0002m /(0.2m * 99ºC)
α = 1.01 *10⁻⁵ (ºC)⁻¹
Answer:
The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.
Explanation:
Given that,
Boiling point = 81.0°C
Atmospheric pressure :
Atmospheric pressure is the force per unit area exerted by the weight of the atmosphere.
The value of atmospheric pressure is

Vapor pressure :
Vapor pressure is equal to the atmospheric pressure.
Hence, The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.
For this problem, we use our knowledge on trigonometric functions on a right triangle to be able to calculate for the x and y components of the walk. We do as follows:
sin 60 = ax / 9.4
ax = 8.14
cos 60 = ay / 9.4
ay = 4.7